Integrand size = 37, antiderivative size = 81 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=-\frac {\log \left (e-2 \sqrt {-d} \sqrt {f} x+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}}+\frac {\log \left (e+2 \sqrt {-d} \sqrt {f} x+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}} \]
-1/4*ln(e+2*f*x^2-2*x*(-d)^(1/2)*f^(1/2))/(-d)^(1/2)/f^(1/2)+1/4*ln(e+2*f* x^2+2*x*(-d)^(1/2)*f^(1/2))/(-d)^(1/2)/f^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(191\) vs. \(2(81)=162\).
Time = 0.09 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.36 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {-\frac {\left (-d-2 e+\sqrt {d} \sqrt {d+2 e}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {d+e-\sqrt {d} \sqrt {d+2 e}}}\right )}{\sqrt {d+e-\sqrt {d} \sqrt {d+2 e}}}-\frac {\left (d+2 e+\sqrt {d} \sqrt {d+2 e}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {d+e+\sqrt {d} \sqrt {d+2 e}}}\right )}{\sqrt {d+e+\sqrt {d} \sqrt {d+2 e}}}}{2 \sqrt {2} \sqrt {d} \sqrt {d+2 e} \sqrt {f}} \]
(-(((-d - 2*e + Sqrt[d]*Sqrt[d + 2*e])*ArcTan[(Sqrt[2]*Sqrt[f]*x)/Sqrt[d + e - Sqrt[d]*Sqrt[d + 2*e]]])/Sqrt[d + e - Sqrt[d]*Sqrt[d + 2*e]]) - ((d + 2*e + Sqrt[d]*Sqrt[d + 2*e])*ArcTan[(Sqrt[2]*Sqrt[f]*x)/Sqrt[d + e + Sqrt [d]*Sqrt[d + 2*e]]])/Sqrt[d + e + Sqrt[d]*Sqrt[d + 2*e]])/(2*Sqrt[2]*Sqrt[ d]*Sqrt[d + 2*e]*Sqrt[f])
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6, 1478, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e-2 f x^2}{4 d f x^2+e^2+4 e f x^2+4 f^2 x^4} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e-2 f x^2}{4 f x^2 (d+e)+e^2+4 f^2 x^4}dx\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle -\frac {\int -\frac {2 \left (\sqrt {-d}-2 \sqrt {f} x\right )}{\sqrt {f} \left (2 x^2-\frac {2 \sqrt {-d} x}{\sqrt {f}}+\frac {e}{f}\right )}dx}{4 \sqrt {-d} \sqrt {f}}-\frac {\int -\frac {2 \left (2 \sqrt {f} x+\sqrt {-d}\right )}{\sqrt {f} \left (2 x^2+\frac {2 \sqrt {-d} x}{\sqrt {f}}+\frac {e}{f}\right )}dx}{4 \sqrt {-d} \sqrt {f}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {-d}-2 \sqrt {f} x}{2 x^2-\frac {2 \sqrt {-d} x}{\sqrt {f}}+\frac {e}{f}}dx}{2 \sqrt {-d} f}+\frac {\int \frac {2 \sqrt {f} x+\sqrt {-d}}{2 x^2+\frac {2 \sqrt {-d} x}{\sqrt {f}}+\frac {e}{f}}dx}{2 \sqrt {-d} f}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\log \left (2 \sqrt {-d} \sqrt {f} x+e+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}}-\frac {\log \left (-2 \sqrt {-d} \sqrt {f} x+e+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}}\) |
-1/4*Log[e - 2*Sqrt[-d]*Sqrt[f]*x + 2*f*x^2]/(Sqrt[-d]*Sqrt[f]) + Log[e + 2*Sqrt[-d]*Sqrt[f]*x + 2*f*x^2]/(4*Sqrt[-d]*Sqrt[f])
3.6.24.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ [c*d^2 - a*e^2, 0] && !GtQ[b^2 - 4*a*c, 0]
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\ln \left (-2 \sqrt {-d f}\, f \,x^{2}-2 d f x -\sqrt {-d f}\, e \right )}{4 \sqrt {-d f}}+\frac {\ln \left (-2 \sqrt {-d f}\, f \,x^{2}+2 d f x -\sqrt {-d f}\, e \right )}{4 \sqrt {-d f}}\) | \(74\) |
default | \(f^{2} \left (-\frac {\left (d f +2 e f -\sqrt {d \,f^{2} \left (d +2 e \right )}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {f x \sqrt {2}}{\sqrt {-d f -e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}\right )}{4 f^{2} \sqrt {d \,f^{2} \left (d +2 e \right )}\, \sqrt {-d f -e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}+\frac {\left (-d f -2 e f -\sqrt {d \,f^{2} \left (d +2 e \right )}\right ) \sqrt {2}\, \arctan \left (\frac {f x \sqrt {2}}{\sqrt {d f +e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}\right )}{4 f^{2} \sqrt {d \,f^{2} \left (d +2 e \right )}\, \sqrt {d f +e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}\right )\) | \(193\) |
-1/4/(-d*f)^(1/2)*ln(-2*(-d*f)^(1/2)*f*x^2-2*d*f*x-(-d*f)^(1/2)*e)+1/4/(-d *f)^(1/2)*ln(-2*(-d*f)^(1/2)*f*x^2+2*d*f*x-(-d*f)^(1/2)*e)
Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.74 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\left [-\frac {\sqrt {-d f} \log \left (\frac {4 \, f^{2} x^{4} - 4 \, {\left (d - e\right )} f x^{2} + e^{2} + 4 \, {\left (2 \, f x^{3} + e x\right )} \sqrt {-d f}}{4 \, f^{2} x^{4} + 4 \, {\left (d + e\right )} f x^{2} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {d f} \arctan \left (\frac {\sqrt {d f} x}{d}\right ) - \sqrt {d f} \arctan \left (\frac {{\left (2 \, f x^{3} + {\left (2 \, d + e\right )} x\right )} \sqrt {d f}}{d e}\right )}{2 \, d f}\right ] \]
[-1/4*sqrt(-d*f)*log((4*f^2*x^4 - 4*(d - e)*f*x^2 + e^2 + 4*(2*f*x^3 + e*x )*sqrt(-d*f))/(4*f^2*x^4 + 4*(d + e)*f*x^2 + e^2))/(d*f), -1/2*(sqrt(d*f)* arctan(sqrt(d*f)*x/d) - sqrt(d*f)*arctan((2*f*x^3 + (2*d + e)*x)*sqrt(d*f) /(d*e)))/(d*f)]
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\sqrt {- \frac {1}{d f}} \log {\left (- d x \sqrt {- \frac {1}{d f}} + \frac {e}{2 f} + x^{2} \right )}}{4} - \frac {\sqrt {- \frac {1}{d f}} \log {\left (d x \sqrt {- \frac {1}{d f}} + \frac {e}{2 f} + x^{2} \right )}}{4} \]
sqrt(-1/(d*f))*log(-d*x*sqrt(-1/(d*f)) + e/(2*f) + x**2)/4 - sqrt(-1/(d*f) )*log(d*x*sqrt(-1/(d*f)) + e/(2*f) + x**2)/4
\[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\int { -\frac {2 \, f x^{2} - e}{4 \, f^{2} x^{4} + 4 \, d f x^{2} + 4 \, e f x^{2} + e^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (61) = 122\).
Time = 0.40 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.28 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\sqrt {2} {\left (d e f {\left | f \right |} - \sqrt {d^{2} + 2 \, d e} {\left (d + e\right )} f^{2} + \sqrt {d^{2} + 2 \, d e} d f^{2}\right )} \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {\frac {4 \, d f + 4 \, e f + \sqrt {-16 \, e^{2} f^{2} + 16 \, {\left (d f + e f\right )}^{2}}}{f^{2}}}}\right )}{4 \, {\left (d^{2} + d e - \sqrt {d^{2} + 2 \, d e} d\right )} \sqrt {{\left (d + e + \sqrt {d^{2} + 2 \, d e}\right )} f} f^{2}} + \frac {\sqrt {2} {\left (d e f {\left | f \right |} + \sqrt {d^{2} + 2 \, d e} {\left (d + e\right )} f^{2} - \sqrt {d^{2} + 2 \, d e} d f^{2}\right )} \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {\frac {4 \, d f + 4 \, e f - \sqrt {-16 \, e^{2} f^{2} + 16 \, {\left (d f + e f\right )}^{2}}}{f^{2}}}}\right )}{4 \, {\left (d^{2} + d e + \sqrt {d^{2} + 2 \, d e} d\right )} \sqrt {{\left (d + e - \sqrt {d^{2} + 2 \, d e}\right )} f} f^{2}} \]
1/4*sqrt(2)*(d*e*f*abs(f) - sqrt(d^2 + 2*d*e)*(d + e)*f^2 + sqrt(d^2 + 2*d *e)*d*f^2)*arctan(4*sqrt(1/2)*x/sqrt((4*d*f + 4*e*f + sqrt(-16*e^2*f^2 + 1 6*(d*f + e*f)^2))/f^2))/((d^2 + d*e - sqrt(d^2 + 2*d*e)*d)*sqrt((d + e + s qrt(d^2 + 2*d*e))*f)*f^2) + 1/4*sqrt(2)*(d*e*f*abs(f) + sqrt(d^2 + 2*d*e)* (d + e)*f^2 - sqrt(d^2 + 2*d*e)*d*f^2)*arctan(4*sqrt(1/2)*x/sqrt((4*d*f + 4*e*f - sqrt(-16*e^2*f^2 + 16*(d*f + e*f)^2))/f^2))/((d^2 + d*e + sqrt(d^2 + 2*d*e)*d)*sqrt((d + e - sqrt(d^2 + 2*d*e))*f)*f^2)
Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\mathrm {atan}\left (\frac {2\,f^{3/2}\,x^3+2\,d\,\sqrt {f}\,x+e\,\sqrt {f}\,x}{\sqrt {d}\,e}\right )-\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {d}}\right )}{2\,\sqrt {d}\,\sqrt {f}} \]